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Solve the following equation$x^{2}r^{2}+2r(2q-p) x+(p-2q)^{2}=0$
Given: $x^{2}r^{2}+2r(2q-p) x+(p-2q)^{2}=0$
To do: Solve the given equation:
Solution:
$x^{2}r^{2}+2r(2q-p) x+(p-2q)^{2}=0$
$(xr)^{2} - 2 \times (xr) \times (p - 2q) + (p - 2q)^{2}= 0$
Using identity $a^{2}- 2ab + b^{2} = (a - b)^{2}$
$(xr - (p - 2q))^{2} = 0$
$(xr - p + 2q)(xr - p + 2q)=0$
or $xr = p - 2q$ or $x = \frac{p - 2q}{r}$ Answer
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